what's Kp if Cl2CO was found to be 7.46% decomposed at 395'C when initial pressure is 7.60atm.
Cl2CO --> <-- CO + Cl2
Initial p = 7.60 atm.
7.46% decomposed.
(CO) = 7.60*0.0746 = ??
(Cl2) = same
((Cl2CO) = 7.60*(0.9254) = ??
Kp = pCO*pCl2/pCl2CO
Post your work if you get stuck. Check my work. Check my thinking.
To determine Kp for the reaction Cl2CO ↔ CO + Cl2, you need to calculate the partial pressures of carbon monoxide (CO), chlorine (Cl2), and chlorine carbonate (Cl2CO) at equilibrium.
Given:
Initial pressure (P₀) = 7.60 atm
Percentage decomposed = 7.46%
First, let's calculate the partial pressures of CO and Cl2 at equilibrium using the percentage decomposed.
Partial pressure of CO:
(CO) = P₀ * (percentage decomposed/100)
= 7.60 atm * (0.0746)
= 0.567 atm
Partial pressure of Cl2:
(Cl2) = P₀ * (percentage decomposed/100)
= 7.60 atm * (0.0746)
= 0.567 atm
Next, calculate the partial pressure of Cl2CO at equilibrium:
Partial pressure of Cl2CO:
(Cl2CO) = P₀ - P(CO) - P(Cl2)
= 7.60 atm - 0.567 atm - 0.567 atm
= 6.466 atm
Now you have the values needed to calculate Kp:
Kp = (P(CO) * P(Cl2)) / P(Cl2CO)
= (0.567 atm * 0.567 atm) / 6.466 atm
= 0.0497
Therefore, Kp for the reaction Cl2CO ↔ CO + Cl2 is approximately 0.0497.
Please note that calculations involving equilibrium constants may be subject to rounding, so always verify with your own work and the given data.